Problem: A cube has side length $6$. Its vertices are alternately colored black and purple, as shown below. What is the volume of the tetrahedron whose corners are the purple vertices of the cube? (A tetrahedron is a pyramid with a triangular base.)

[asy]
import three;
real t=-0.05;
triple A,B,C,D,EE,F,G,H;
A = (0,0,0);
B = (cos(t),sin(t),0);
D= (-sin(t),cos(t),0);
C = B+D;
EE = (0,0,1);
F = B+EE;
G = C + EE;
H = D + EE;
draw(surface(B--EE--G--cycle),rgb(.6,.3,.6),nolight);
draw(surface(B--D--G--cycle),rgb(.7,.4,.7),nolight);
draw(surface(D--EE--G--cycle),rgb(.8,.5,.8),nolight);
draw(B--C--D);
draw(EE--F--G--H--EE);
draw(B--F);
draw(C--G);
draw(D--H);
pen pu=rgb(.5,.2,.5)+8; pen bk=black+8;
dot(B,pu); dot(C,bk); dot(D,pu); dot(EE,pu); dot(F,bk); dot(G,pu); dot(H,bk);
[/asy]
Solution: The volume of any pyramid is $\frac 13$ the product of the base area and the height. However, determining the height of the purple tetrahedron is somewhat tricky! Instead of doing that, we observe that the total volume of the cube consists of the purple tetrahedron and four other "clear" tetrahedra. Each clear tetrahedron is formed by one of the black vertices of the cube together with its three purple neighbors. The clear tetrahedra are convenient to work with because they have lots of right angles.

Each clear tetrahedron has an isosceles right triangular base of area $\frac 12\cdot 6\cdot 6 = 18$, with corresponding height $6$ (a side of the cube). Thus, each clear tetrahedron has volume $\frac 13\cdot 18\cdot 6 = 36$.

The cube has volume $6^3 = 216$. The volume of the purple tetrahedron is equal to the volume of the cube minus the volume of the four clear tetrahedra. This is $216 - 4\cdot 36 = \boxed{72}$.